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For the interaction between electromagnetic fields and matter,

  1. when do we have to include quantization of the EM field and when we can ignore it?
  2. when do we have to include quantization of atomic energy levels and when we can ignore it?

Update: I am aware that part of the answer might depend on the accuracy we are looking for. Part of the problem here is that I do not know how to estimate such things, or what quantity will quantify the accuracy we are looking for which can tell us if we can ignore quantization of either energy levels or the fields or not.

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Who would like to write another textbook on spectroscopy? –  Georg Dec 2 '11 at 11:10
    
@Georg Could you please point out a reference? –  Revo Dec 2 '11 at 17:22
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@Georg Since obviously your are knowledgeable person, yes I was asking you to recommend a textbook. Googling a keyword will not give me the best book that addresses my question. But you know what, on a second thought I do not want anything from someone who communicates with others with such an arrogant tone! –  Revo Dec 2 '11 at 19:51
    
I am 65 years now, my textbooks are totally outdated. –  Georg Dec 2 '11 at 20:15
    
A good (but expensive) reference is Demtroeder. –  Antillar Maximus Dec 2 '11 at 21:32

4 Answers 4

I'm not sure there is a generic answer to your questions other than the trivial "don't bother including the quantization when the accuracy of your result isn't compromised by making this approximation". I know that doesn't really help much, because you may not be able to verify this until you've done the calculation including the quantization anyway. You may have no choice in the matter - modelling everything at a microscopic level may just be intractable.

Sometimes the answers as to when you have to do the full quantum calculation are surprising. For example, it is a common belief that explanation of the photoelectric effect requires you to treat the electromagnetic field quantum mechanically - i.e. you need photons. However, computations in Mandel and Wolf reveal that the observed experimental outcome can be obtained by treating the radiation purely classically (but the atomic electrons quantum mechanically).

(Of course light does have a quantum nature, as revealed by photon antibunching).

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the same is true for the Compton effect, btw - a semi-classical explanation was given by Schrödinger in 1927 –  Christoph Dec 2 '11 at 12:23
    
@Christoph: Thanks, I didn't know that about the Compton effect... –  twistor59 Dec 2 '11 at 12:36
    
Photoelectric effect can also be described without photons according to the seminal paper by Lamb and Scully. ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/… –  Revo Dec 2 '11 at 19:56
    
@twistor59: This answer is only technically correct. Mandel and Wolf are saying that if you send a coherent wave with enormous occupation numbers, so that it can be treated as a classical wave, you get excitation of the atoms by (semiclassical) single-photon absorption only when the frequency is right to excite the atoms. But if you explain the photoelectric effect semiclassically, you can't account for the fact that each absorbed photon is removed from the EM wave. The reradiated wave from the atom will cancel part of the semiclassical wave and give statistical absoprtion of photons... –  Ron Maimon Dec 2 '11 at 20:15
    
But the reradiated wave goes outward at the speed of light, and cannot reduce the classical wave amplitude instantaneously, it doesn't collapse the photon wavefunction. This means that if you have only a few photons, they can be simultaneously absorbed at far away places, and more absoprtion events can happen semiclassically than there are photons to produce the events. This leads to violations of energy conservation. This was Bohr's motivation for BKS, and Kramers and Slater joined him in abandoning energy conservation. This is not correct, however, energy is conserved. –  Ron Maimon Dec 2 '11 at 20:16

Generally speaking, the answer to both questions is linked to some number becoming increasingly large so that, for atoms you have a large density of higher excited states (think to Rydberg atoms as an example) or for electromagnetic field one has such a large number of photons that a coherent state is a good description of it and an average field can be safely taken. Then, quantum fluctuations are negligible small as these numbers increases.

In order to make all the argument quantitative, let me consider a standard Hamiltonian for radiation-matter interaction for hydrogen-like atoms, in a non-relativistic limit,

$$H=H_a+H_f+H_i=H_f-\frac{\hbar^2}{2m}\Delta_2-\frac{Ze^2}{r}-{\bf d}\cdot{\bf E}$$

where I have used an equivalent form for the interaction. Now, we can always rewrite this through a complete set of atomic states and this will give (the continuous part of the spectrum is implicit in the summation)

$$H=H_f+\sum_nE_n|n\rangle\langle n|+\sum_{m,n}|m\rangle\langle n|{\bf d}_{mn}\cdot{\bf E}$$

but we can do the same also for the field. Assuming this monochromatic and using coherent states $|\alpha\rangle$, that we know are overcomplete $\langle\alpha|\beta\rangle=$, we can use the resolution of identity (e.g. see here)

$$I=\int\frac{d^2\alpha}{\pi}|\alpha\rangle\langle\alpha|$$

that will produce

$$H=H_f+\sum_nE_n|n\rangle\langle n|+\sum_{m,n}|m\rangle\langle n|{\bf d}_{mn}\cdot\int \frac{d^2\alpha}{\pi}\frac{d^2\beta}{\pi}\langle\alpha|{\bf E}|\beta\rangle|\alpha\rangle\langle\beta|.$$

Now, we are a step away from our conclusion. Indeed, we should not that $|\alpha|^2=N$, being $N$ the number of photons. So, the interaction part of the Hamiltonian can be promptly evaluated as

$$\langle\alpha|{\bf E}|\beta\rangle=\tilde{\bf E}(\alpha,\beta)e^{-\frac{1}{2}|\alpha-\beta|^2}$$

and, for a very large $N$, the integral will have a dominant contribution and the coherent state can be assumed practically orthogonal. This will justify the use of a classical approximation through the averaged field.

This argument can be repeated for the atomic states, if we introduce the operators (see here) $\sigma_{nm}=|n\rangle\langle m|$, $\sigma_{nm}^\dagger=|n\rangle\langle m|$ and $\sigma_{nm}^3=(1/2)(|n\rangle\langle n|-|m\rangle\langle m|$ forming an su(2) algebra. We can use now atomic coherent states and arrive to the same conclusion as above, provided the atomic state is large enough. This is the rationale behind this kind of approximations normally used in quantum optics.

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In general, a semi-classical approach is easier to handle mathematically and is taught before attacking the full quantum description. I don't know if there is are any hard and fast rules on what approach to use, but having experimental data would be the best way to check. I don't know about other areas, but in nonlinear optics when the light fields are intense, a semi-classical approach works just as well as a fully quantum approach. (Semi-classical: Atomic part is quantized while the Optical part is a classical wave). In many cases, it is relatively straightforward to go from the semi-classical picture to the quantum picture (ex: replace a classical EM Field with a coherent state).

There are many cases where a semi-classical approach fails and then you have no choice but to recast the problem using a fully quantum picture. To summarize, it is all about convenience and what you are after. A fully quantum picture is definitely richer, but may not be necessary.

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This question has been a personal crusade of mine for the last 20 years. Around that time I figured out that the usual explanation for the photo-electric effect was based on basic misconceptions about the way waves transferred energy. To put things in a nutshell, if the usual power density argument were taken literally, then not only the photo-electric effect but such commonplace devices as the crystal radio should be "classically impossible". It was almost ten years later that I figured out the explanation for the Compton effect based on standing waves, and I was hugely disappointed when I learned that Schroedinger had already published my explanation in 1927.

Since then I have put together two more pieces of the puzzle. The Black Body Spectrum was a tough one until I figured out how to calculate the equilibrium between a mechanical oscillator and the radiation field. Then I was able to show that the classical radiation field had to follow the natural equilibrium of a system of mechanical oscillators, and not vice versa. If the equipartition theorem failed at the mechanical level (which it does), then you don't need to quantize the electormagnetic field to get the Planck distribution: it follows automatically from the mechanical equilibrium. I explain this in a series of blogposts beginning with this one in my blog "Why I Hate Physics".

The other big problem I solved was how to explain the very baffling question of the flecks of silver appearing on a photographic plate when exposed to the light of a distant star. According to classical theory the energy of light can be made arbitrarily diffuse. How then can it accumulate with sufficient intensity to provide the significant amount of energy needed to drive the chemical reduction of Ag+ to metallic silver in the silver bromide crystal? I was able to explain this by showing that the energy for the transition is already available in the detector system, namely the silver bromide crystal. It is not enough to look at the enthalpy of the transition; you must really consider the Gibbs Free Energy, which includes a term for the entropy. Treating the crystal as a solid solution of metallic silver and silver bromide, it is easily shown that at the very low concentrations present in a photographic film, the chemical transition is very nearly spontaneous. So no bunching of energy in the form of photons is required.

I also demonstrate a mechanism whereby the energy of the crystal is concentrated at the point of detection. I call this phenomenon Quantum Siphoning and it is explained in the linked article.

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So you think you can explain silver film, how about all of the other photon-counting detectors we have today? I didn't bother to read your article btw. –  user2963 Dec 2 '11 at 14:21
    
The silver film is the toughest because unlike all your other photon-counting detectors, you don't plug it into the wall. So there's no obvious source for the energy other than the incoming light. –  Marty Green Dec 2 '11 at 14:27
    
The silver film is not more difficult than any other photo receptor! You always have the old age (meanwhile) problem that You have to apply photon nature of light in this case, or wave nature for eg diffraction of light for easy understanding. The problem is that "tertium non datur". This is the core problem of QM and human thoughts. The problems Marty invents on silver halide process origin from his not knowing of the latent images nature. He thinks in wrong simple chemical ways. The latent image is hard core crystal physics. –  Georg Dec 2 '11 at 14:52
    
How do you explain the Lamb shift, then? –  Jerry Schirmer Dec 2 '11 at 16:51
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Instead of focusing on this answer, consider the case of two entangled photons with opposite polarization. If you see one with one polarization, the other has the opposite polarization. How do you describe this classically? The EM wave has one polarization only, it cannot have a polarization that is correlated with another far away. Worse, the correlations in the polarizations are observed by Aspect et al and they violate the Bell inequality. This conclusively demonstrates you need quantum states of superposition for the electromagnetic field, ignoring photoelectric effect or atomic stuff. –  Ron Maimon Dec 3 '11 at 6:42

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